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The Pythagorean trigonometric identity is a trigonometric identity expressing the Pythagorean theorem in terms of trigonometric functions. Along with the sum-of-angles formulae it is the basic relation among the sin and cos functions from which all others may be derived.

Statement of the identity
Mathematically, the Pythagorean identity states:


 * $$\sin^2 x + \cos^2 x = 1.\qquad\qquad (1) \!$$

(Note that sin2 x means (sin x)2.)

Two more Pythagorean trigonometric identities can be identified. They are derived as follows.
 * $$\begin{align}

\frac{\sin^2 x}{\sin^2 x} + \frac{\cos^2 x}{\sin^2 x}&=\frac{1}{\sin^2 x} & \qquad\frac{\sin^2 x}{\cos^2 x} + \frac{\cos^2 x}{\cos^2 x}&=\frac{1}{\cos^2 x}\\ 1 + \cot^2 x &=\csc^2 x & \tan^2 x + 1 &=\sec^2 x \end{align}$$ Like (1), they also have simple geometric interpretations as instances of the Pythagorean theorem.

Using right-angled triangles
Using the elementary "definition" of the trigonometric functions in terms of the sides of a right triangle,


 * $$\sin x = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}}$$
 * $$\cos x = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$$

the theorem follows by squaring both and adding; the left-hand side of the identity then becomes


 * $$\frac{\mathrm{opposite}^2 + \mathrm{adjacent}^2}{\mathrm{hypotenuse}^2}$$

which by the Pythagorean theorem is equal to 1. Note, however, that this definition is only valid for angles between 0 and &frac12;&pi; radians (not inclusive) and therefore this argument does not prove the identity for all angles. Values of 0 and &frac12;&pi; are trivially proven by direct evaluation of sin and cos at those angles.

To complete the proof, the identities of Trigonometric symmetry, shifts, and periodicity must be employed. By the periodicity identities we can say if the formula is true for -&pi; < x ≤ &pi; then it is true for all real x. Next we prove the range &frac12;&pi; < x ≤ &pi;, to do this we let t = x - &frac12;&pi;, t will now be in the range 0 < x ≤ &frac12;&pi;. We can then make use of squared versions of some basic shift identites (squaring conveniently removes the minus signs).


 * $$\sin^2x+\cos^2x \equiv \sin^2\left(t+\frac{1}{2}\pi\right) + \cos^2\left(t+\frac{1}{2}\pi\right) \equiv \cos^2t+\sin^2t \equiv 1.$$

All that remains is to prove it for &minus;&pi; < x < 0; this can be done by squaring the symmetry identities to get


 * $$\sin^2x\equiv\sin^2(-x)\mbox{ and }\cos^2x\equiv\cos^2(-x)\,.$$

Using the unit circle
If the trigonometric functions are defined in terms of the unit circle, the proof is immediate: given an angle θ, there is a unique point P on the unit circle centered at the origin in the Euclidean plane at an angle θ from the x-axis, and cos θ, sin θ are respectively the x- and y-coordinates of P. By definition of the unit circle, the sum of the squares of these coordinates is 1, whence the identity.

The relationship to the Pythagorean theorem is through the fact that the unit circle is actually defined by the equation


 * $$x^2 + y^2 = 1\,.$$

Since the x- and y-axes are perpendicular, this fact is actually equivalent to the Pythagorean theorem for triangles with hypotenuse of length 1 (which is in turn equivalent to the full Pythagorean theorem by applying a similar-triangles argument).

Using power series
The trigonometric functions may also be defined using power series, namely (for x an angle measured in radians):


 * $$\sin x = \sum_{n = 0}^\infty \frac{(-1)^n}{(2n + 1)!} x^{2n + 1}$$
 * $$\cos x = \sum_{n = 0}^\infty \frac{(-1)^n}{(2n)!} x^{2n}$$

Using the formal multiplication law for power series we obtain




 * $$\sin^2 x\, $$
 * $$= \sum_{i = 0}^\infty \sum_{j = 0}^\infty \frac{(-1)^i}{(2i + 1)!} \frac{(-1)^j}{(2j + 1)!} x^{(2i + 1) + (2j + 1)}$$
 * $$= \sum_{n = 1}^\infty \left(\sum_{i = 0}^{n - 1} \frac{(-1)^{n - 1}}{(2i + 1)!(2(n - i - 1) + 1)!}\right) x^{2n}$$
 * $$= \sum_{n = 1}^\infty \left( \sum_{i = 0}^{n - 1} {2n \choose 2i + 1} \right) \frac{(-1)^{n - 1}}{(2n)!} x^{2n}$$
 * }
 * $$= \sum_{n = 1}^\infty \left(\sum_{i = 0}^{n - 1} \frac{(-1)^{n - 1}}{(2i + 1)!(2(n - i - 1) + 1)!}\right) x^{2n}$$
 * $$= \sum_{n = 1}^\infty \left( \sum_{i = 0}^{n - 1} {2n \choose 2i + 1} \right) \frac{(-1)^{n - 1}}{(2n)!} x^{2n}$$
 * }
 * $$= \sum_{n = 1}^\infty \left( \sum_{i = 0}^{n - 1} {2n \choose 2i + 1} \right) \frac{(-1)^{n - 1}}{(2n)!} x^{2n}$$
 * }


 * $$\cos^2 x\, $$
 * $$= \sum_{i = 0}^\infty \sum_{j = 0}^\infty \frac{(-1)^i}{(2i)!} \frac{(-1)^j}{(2j)!} x^{(2i) + (2j)}$$
 * $$= \sum_{n = 0}^\infty \left(\sum_{i = 0}^n \frac{(-1)^n}{(2i)!(2(n - i))!}\right) x^{2n}$$
 * $$= \sum_{n = 0}^\infty \left( \sum_{i = 0}^n {2n \choose 2i} \right) \frac{(-1)^n}{(2n)!} x^{2n} $$
 * }
 * $$= \sum_{n = 0}^\infty \left( \sum_{i = 0}^n {2n \choose 2i} \right) \frac{(-1)^n}{(2n)!} x^{2n} $$
 * }
 * $$= \sum_{n = 0}^\infty \left( \sum_{i = 0}^n {2n \choose 2i} \right) \frac{(-1)^n}{(2n)!} x^{2n} $$
 * }

Note that in the expression for $$\sin^2$$, n must be at least 1, while in the expression for $$\cos^2$$, the constant term is equal to 1. The remaining terms of their sum are (with common factors removed)


 * $$\sum_{i = 0}^n {2n \choose 2i} - \sum_{i = 0}^{n - 1} {2n \choose 2i + 1}

= \sum_{j = 0}^{2n} (-1)^j {2n \choose j} = (1 - 1)^{2n} = 0 $$

by the binomial theorem. The Pythagorean theorem is not closely related to the Pythagorean identity when the trigonometric functions are defined in this way; instead, in combination with the theorem, the identity now shows that these power series parameterize the unit circle, which we used in the previous section. Note that this definition actually constructs the sin and cos functions in a rigorous fashion and proves that they are differentiable, so that in fact it subsumes the previous two.

Using the differential equation
It is possible to define the sin and cos functions as the two unique solutions to the differential equation


 * $$y'' + y = 0$$

satisfying respectively $$y(0) = 0, y'(0) = 1$$ and $$y(0) = 1, y'(0) = 0$$. It follows from the theory of ordinary differential equations that the former solution, sin, has the latter, cos, as its derivative, and it follows from this that the derivative of cos is &minus;sin. To prove the Pythagorean identity it suffices to show that the function


 * $$z = \sin^2 x + \cos^2 x$$

is constant and equal to 1. However, differentiating it and applying the two facts just mentioned we see that $$z' = 0$$ so z is constant, and $$z(0) = 1$$.

This form of the identity likewise has no direct connection with the Pythagorean theorem.

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